If red light is replaced by blue light in single slit diffraction then

The correct answer is Option (2) → diffraction bands become narrower

In single slit diffraction the minima lie at \( a\sin\theta = n\lambda \), so for a slit of fixed width a the angular position of every band is proportional to the wavelength.

    \( \displaystyle \theta \approx \frac{n\lambda}{a} \quad (\text{small angles}) \)

Blue light has a shorter wavelength than red light (about 450 nm against 700 nm), so \( \theta \) becomes smaller for every band. The whole pattern, including the central maximum, shrinks and the bands are closer together, that is, narrower. The pattern does not vanish, because diffraction occurs for any wavelength.

NCERT (2026-27):
The Single Slit – “… It is observed that the intensity has a central maximum at θ = 0 and other secondary maxima at \( \theta \simeq (n+1/2)\lambda/a \), which go on becoming weaker and weaker with increasing n. The minima (zero intensity) are at \( \theta \simeq n\lambda/a \), n = ±1, ±2, ±3, ....”